Evaluate the indefinite integral?

Question

(x+21) times the square root of 42x+x^2

i set u to 42x+x^2

and du to 42+2x dx

but i keep getting the wrong answer after substituting it back in

can you please give me the answer i came up with 4 different ones, thank you

Answer 1

∫ (x + 21) √(42x + x²) dx =

you set u-substitution correctly:

since your integrand includes both the radicand (42x + x²) and its half-derivative (x + 21)

let 42x + x² = u

differentiate both sides:

d(42x + x²) = du →

(42 + 2x) dx = du →

2(21 + x) dx = du →

2(21 + x) dx = (1/2)du

then, substituting, you get:

∫ √(42x + x²) (21 + x) dx = ∫ √u (1/2)du =

taking out the constant and expressing the integrand as a fractionary power,

(1/2) ∫ u^(1/2) du =

(1/2) {u^[(1/2) + 1]} /[(1/2) + 1] + C =

(1/2) [u^(3/2)] /(3/2) + C =

(1/2)(2/3) [u^(3/2)] + C =

(1/3) √u³ + C = (1/3) u√u + C

that is, substituting back u = (42x + x²)

∫ (x + 21) √(42x + x²) dx = (1/3) (42x+ x²)√(42x+ x²) + C

I hope it helps...

Bye!

Answer 2

Let S be the symbol of integral.

S (x+21).sqrt(42x + x^2) dx

Subtitute 42x + x^2 = u

Then (42 + 2x) dx = du

(x+21)dx = 0.5 du

S sqrt(42x + x^2) . (x+21)dx = S sqrt(u) . du

= S u^(1/2) du = (2/3) . u^(3/2) + c

Subtituting back in...

(2/3) . (42x + x^2)^(3/2) + c

that is the answer.